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1Mechanics of Solids I Mechanics of Solids IAnalysis and Design of Beams for BendingIntroductionIntroduction2o Members that are slender and support loadings appliedperpendicular to their longitudinal axis are calledbeamsIntroductionIntroductionperpendicular to their longitudinal axis are calledbeamso Transverse loadings of beams are classifiedIntroductionIntroductionas concentrated loads or distributed loads.o Applied loads result in internal forcesconsisting of a shear force (from the shearstress distribution) and a bending couple(from the normal stress distribution).o Normal stress is often the critical design criteriax mMc MMyI I Sσ σ= − = =o Requires determination of the location andmagnitude of largest bending moment.3Shear and Bending Moment Diagramso Determination of maximum normal andshearing stresses requires identification ofmaximum internal shear force and bendingcouple.o passing a section through the beam andapplying an equilibrium analysis.o Shear and bending-moment functions mustbe determined for each region of the beambetween any two discontinuities of loadingShear and Bending Moment Diagramso Sign conventions:g4Example Example 55..11For the timber beam and loadingshown, draw the shear and bendingmoment diagrams and determine themaximum normal stress due to bending.Relations among Load, Shear, and Bending Relations among Load, Shear, and BendingMomentMomento Relationship between load and shear:()〺 0yF V V V w xV w x=− + Δ − Δ =Δ = − Δ∑dVwdx= −slope of SFD = - wDCxD CxV V w dx− = −∫shear difference = - area under loading5Relations among Load, Shear, and Bending Relations among Load, Shear, and BendingMomentMomentoRelationship between shear and bending moment:( )( )2120:02CxM M M M V x w xM V x w x′Δ=+ Δ − − Δ + Δ =Δ = Δ − Δ∑dMV=oRelationship between shear and bending moment:slope of BMD = shear at thepointDCxD CxVdxMM Vdx− =∫p parea under SFD = moment differenceRelations among Load, Shear, and BendingRelations among Load, Shear, and BendingMomentMomentoExample: Draw SFD and BMD for simplysupported beams as shownoExample: Draw SFD and BMD for simply-supported beams as shownLwPa2aCa2aCM6Example Example 55..22Draw the shear and bending-momentdiagrams for the beam and loading showndiagrams for the beam and loading shown.ExampleExample 55..33The structure shown is constructed of a W250 ×167 rolled-steel beam. (a) Draw the shear andbending-moment diagrams for the beam andthe given loading (b) Determine normal stressthe given loading.(b) Determine normal stressin sections just to the right and left of point D.76.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEARAND MOMENT DIAGRAMSDesign of Beams for BendingDesign of Beams for Bendingo The largest normal stress is found at the surface where the maximumbending moment occurs.max maxmM c MI Sσ = =o A safe design requires that the maximum normal stress be less thanthe allowable stress for the material used.maxm allMSσ σ≤o Among beam section choices which have an acceptable sectionmodulus, the one with the smallest weight per unit length or cross-sectional area will be the least expensive and the best choice.maxminallSσ=8Example Example 55..44A simply supported steel beam is to carrythe distributed and concentrated loadsthe distributed and concentrated loadsshown. Knowing that the allowable normalstress for the grade of steel to be used is 160MPa, select the wide-flange shape thatshould be used.9Principle Stresses in a BeamPrinciple Stresses in a Beam• Prismatic beam subjected to transverseloadingloadingσ στ τ= − == − =x mxy mMy McI IVQ VQIt It• Determine if the maximum normal stresswithin the cross-section is larger thanσ =mMcI10Principle Stresses in a BeamPrinciple Stresses in a Beam• Cross-section shape results in large values ofnear the s rface hereis also largeτxynear the surfacewhereσxis also large.• σmaxmay be greater than σm.11Example Example 88..11A 160-kN force is applied at the endof a W200 × 52 rolled-steel beam.Neglecting the effects of fillets and ofstress concentrations, determinewhether the normal stresses satisfy adesign specification that they beequal to or less than 150 MPa atsection A−A’.ExampleExample 88..22The overhanging beam supports auniformlydistributed load and ayconcentrated load. Knowing thatfor the grade of steel to be used σall= 165 MPa and τall= 100 MPa, selectthe wide-flange beam whichshould be used.